# 1352.Product-of-the-Last-K-Numbers

## ้ข็ฎๅฐๅ

https://leetcode.com/problems/product-of-the-last-k-numbers/

## ้ข็ฎๆ่ฟฐ

``````Implement the class ProductOfNumbers that supports two methods:

Adds the number num to the back of the current list of numbers.
2. getProduct(int k)

Returns the product of the last k numbers in the current list.
You can assume that always the current list has at least k numbers.
At any time, the product of any contiguous sequence of numbers will fit into a single 32-bit integer without overflowing.

Example:
Input
[[],[3],[0],[2],[5],[4],[2],[3],[4],[8],[2]]

Output
[null,null,null,null,null,null,20,40,0,null,32]

Explanation
ProductOfNumbers productOfNumbers = new ProductOfNumbers();
productOfNumbers.getProduct(2); // return 20. The product of the last 2 numbers is 5 * 4 = 20
productOfNumbers.getProduct(3); // return 40. The product of the last 3 numbers is 2 * 5 * 4 = 40
productOfNumbers.getProduct(4); // return 0. The product of the last 4 numbers is 0 * 2 * 5 * 4 = 0
productOfNumbers.getProduct(2); // return 32. The product of the last 2 numbers is 4 * 8 = 32

Constraints:
There will be at most 40000 operations considering both add and getProduct.
0 <= num <= 100
1 <= k <= 40000``````

## ไปฃ็ 

### Approach #1

``````class ProductOfNumbers {

List<Integer> prod;
int p = 1;
public ProductOfNumbers() {
prod = new ArrayList();
}

public void add(int num) {
if (num == 0) {
p = 1;
prod = new ArrayList();
} else {
p *= num;
}
}

public int getProduct(int k) {
if (k >= prod.size()) {
return 0;
}

int last = prod.get(prod.size() - 1);
int prev = prod.get(prod.size() - 1 - k);
return last / prev;
}
}

/**
* Your ProductOfNumbers object will be instantiated and called as such:
* ProductOfNumbers obj = new ProductOfNumbers();